Math 360, Fall 2014, Assignment 3
By one of those caprices of the mind, which we are perhaps most subject to in early youth, I at once gave up my former occupations; set down natural history and all its progeny as a deformed and abortive creation; and entertained the greatest disdain for a would-be science, which could never even step within the threshold of real knowledge. In this mood of mind I betook myself to the mathematics, and the branches of study appertaining to that science, as being built upon secure foundations, and so, worthy of my consideration.
- - Mary Shelley, Frankenstein
Carefully define the following terms, then give one example and one non-example of each:[edit]
- Isomorphism (from one binary structure to another).
- Isomorphic (binary structures).
- Structural property (of a binary structure).
- Identity element.
- Inverse (of an element of some binary structure with an identity).
- Group.
- Abelian group.
Carefully state the following theorems (you do not need to prove them):[edit]
- Theorem concerning the uniqueness of identity elements (Theorem 3.13).
- Theorem concerning the uniqueness of inverses in groups (second part of Theorem 4.17).
Solve the following problems:[edit]
- Section 3, problems 2, 4, 8, 9, and 17.
- Section 4, problems 1, 5, 7, 12, 13, and 17.
Questions:[edit]
Solutions:[edit]
In an effort to promote collaboration, I will only post work on problems whose problem numbers are elements of $[2]_{3}$. Please join me in working on the problems here. -Ian
Definitions:[edit]
- A structural property of a binary structure $(S,\ast)$ is a property that holds for all structures that are isomorphic to $(S,\ast)$.
- A group $(G,\ast)$ is a binary structure for which the following three properties hold:
- $\ast$ is associative
- $G$ contains an identity element $e$, i.e. for all $a\in G$, $a\ast e = e\ast a = a$
- For all $a\in G$, there exists $a'\in G$ with $a\ast a'=a'\ast a=e$
Theorems:[edit]
- In any group $(G,\ast)$, for each $a\in G$, there is only one element $a'\in G$ with $a\ast a'=a'\ast a=e$, where $e$ is the identity element.
Problems:[edit]
Section 3[edit]
Section 4[edit]
1)Not a group. $(\mathbb{Z},\ast)$ fails to have an inverse at $0$.
5)Not a group. Ordinary division is not associative.
7)$(\mathbb{Z}_{1000},+_{1000})$
12)Not a group. Any matrices with a zero along the diagonal fail to have an inverse.
13)Is a group. The multiplication of two $n \times n$ diagonal matrices with nonzero diagonal entries yields an $n \times n$ diagonal matrix with nonzero diagonal entries, matrix multiplication is associative, the Identity Matrix is an identity, and any such matrices will have non-zero determinants, meaning they will have inverses (diagonal with nonzero diagonal entries.(.
17)Is a group. Multiplying two $n \times n$ upper-triangular matrices with determinant 1 gives another $n \times n$ upper-triangular matrix with determinant 1. Matrix multiplication is associative. The Identity Matrix is upper-trianguler and has determinant one, and the inverse of an $n \times n$ upper-triangular matrix with determinant 1 will also be an $n \times n$ upper-triangular matrix with determinant 1.